\newproblem{lay:5_2_18}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.2.18}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	It can be shown that the algebraic multiplicity of an eigenvalue $\lambda$ is always greater than or equal to the dimension of the eigenspace corresponding to
	$\lambda$. Find $h$ in the matrix $A$ below such that the eigenspace of $\lambda=4$ is two dimensional.
	\begin{center}
		$A=\begin{pmatrix}4 & 2 & 3 & 3 \\ 0 & 2 & h & 3 \\ 0 & 0& 4 & 14 \\ 0 & 0 & 0 & 2\end{pmatrix}$
	\end{center}
}{
   % Solution
	Let's calculate the eigenspace associated to the eigenvalue $\lambda=4$. For doing so we solve the homogenous equation $(A-\lambda I)\mathbf{x}=\mathbf{0}$ making
	use of the augmented matrix below
	\begin{center}
		$\left(\begin{array}{rrrr|r}0 & 2 & 3 & 3 & 0\\ 0 & -2 & h & 3 &0\\ 0 & 0& 0 & 14 &0\\ 0 & 0 & 0 & -2 &0\end{array}\right)\sim
		 \left(\begin{array}{rrrr|r}0 & 1 & \frac{3}{2} & 0 & 0\\ 0 & 0 & h+3 & 0 &0\\ 0 & 0& 0 & 1 &0\\ 0 & 0 & 0 & 0 &0\end{array}\right)$
	\end{center}
	Note that we have made the elements $a_{13}=0$ and $a_{23}=1$ because for doing that we would need to perform the row operations
	\begin{center}
		$\mathbf{r}_1\leftarrow \mathbf{r}_1 -\frac{\frac{3}{2}}{h+3}\mathbf{r}_2$\\
		$\mathbf{r}_2\leftarrow \frac{1}{h+3}\mathbf{r}_2$
	\end{center}
	which are not allowed if $h=-3$. If $h\neq -3$, then the eigenspace is formed by all the vectors of the form $\{x_1,0,0,0\}$ whose dimension is 1. If
	$h=-3$, then the eigenspace is formed by all the vectors of the form $\{x_1,-\frac{3}{2}x_3,x_3,0\}$ whose dimension is 2.
	
}
\useproblem{lay:5_2_18}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
